askiitians.com0Home » Forum » Electrostatics » Two balls of charges q1 and q2 initially...Two balls of charges q1 and q2 initially have the exact same velocity. Both the balls are subjected to same uniform electric field for the same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by 60°. The direction of velocity of second ball is found to change by 90°. (i) The electric field and the initial velocity of the charged particle are inclined at what angle? (ii) The new velocity of the second charged particle has a magnitude `x` times the initial velocity. What is the value of x? (iii) If the specific charge ( charge to mass ratio) of the first ball is k, what is the specific charge of the second ball ?

Let v1 and v2 be the velocities of the first and second balls after the removal of the uniform electric field. By hypothesis, the angle between the velocity v1 and the initial velocity v is 60°. Therefore, the change in the momentum of the first ball is

Δp1 = q1 E = Δt = m1v sin60

Here we use the condition that v1 = v/2, which implies that the change in the momentum Δp1 of the first ball occurs in a direction perpendicular to the direction of its velocity v1

Since E || Δp1 and the direction of variation of the second ball momentum is parallel to the direction of Δp1, we obtain for the velocity of the second ball (it can easily be seen that the charges on the balls have the same sign)

V2 = vt an30 = v/√3

The corresponding change in the momentum of the second ball is

Δp2 = q2 E Δt = m2V/cos 30

Hence we obtain

q1/q2 = m1 sin60/m2/cos 30

q2/m2 = 4/3 * q1/m1 = 4/4 k1